POJ 1789 Truck History 解题报告
分类：图论，生成树，Prim
作者：
ACShiryu
时间：2011-7-29
Truck History
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 9712 | Accepted: 3579 |
Description
Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for bricks. The company has its own code describing each type of a truck. The code is simply a string of exactly seven lowercase letters (each letter on each position has a very special meaning but that is unimportant for this task). At the beginning of company's history, just a single truck type was used but later other types were derived from it, then from the new types another types were derived, and so on.
Today, ACM is rich enough to pay historians to study its history. One thing historians tried to find out is so called derivation plan -- i.e. how the truck types were derived. They defined the distance of truck types as the number of positions with different letters in truck type codes. They also assumed that each truck type was derived from exactly one other truck type (except for the first truck type which was not derived from any other type). The quality of a derivation plan was then defined as
1/Σ(to,td)d(to,td)
where the sum goes over all pairs of types in the derivation plan such that t o is the original type and t d the type derived from it and d(t o,t d) is the distance of the types.
Since historians failed, you are to write a program to help them. Given the codes of truck types, your program should find the highest possible quality of a derivation plan.
Today, ACM is rich enough to pay historians to study its history. One thing historians tried to find out is so called derivation plan -- i.e. how the truck types were derived. They defined the distance of truck types as the number of positions with different letters in truck type codes. They also assumed that each truck type was derived from exactly one other truck type (except for the first truck type which was not derived from any other type). The quality of a derivation plan was then defined as
where the sum goes over all pairs of types in the derivation plan such that t o is the original type and t d the type derived from it and d(t o,t d) is the distance of the types.
Since historians failed, you are to write a program to help them. Given the codes of truck types, your program should find the highest possible quality of a derivation plan.
Input
The input consists of several test cases. Each test case begins with a line containing the number of truck types, N, 2 <= N <= 2 000. Each of the following N lines of input contains one truck type code (a string of seven lowercase letters). You may assume that the codes uniquely describe the trucks, i.e., no two of these N lines are the same. The input is terminated with zero at the place of number of truck types.
Output
For each test case, your program should output the text "The highest possible quality is 1/Q.", where 1/Q is the quality of the best derivation plan.
Sample Input
4 aaaaaaa baaaaaa abaaaaa aabaaaa 0
Sample Output
The highest possible quality is 1/3.
题目大意是就是给出n个长度为7的字符串，每个字符串代表一个车，定义车的距离是两个字符串间不同字母的个数，题目要求的数不同的车的距离的最小值，即所求的就是最小生成树
关于数据的输入和输出详见样例，要注意输出完数据后还有个'.'，这题是一个稠密图，用Prim算法比较好，关于Prim算法详见：
http://www.nocow.cn/index.php/Prim%E7%AE%97%E6%B3%95
这题提交一次就过了，算是比较基础的生成树
参考代码：
1 #include<iostream>
2 #include<cstdlib>
3 #include<cstdio>
4 #include<cstring>
5 #include<algorithm>
6 #include<cmath>
7 using namespace std;
8 char str[2000][8];
9 int dis[2000][2000];
10 int lowdis[2000];
11 int main()
12 {
13 int n ;
14 while ( cin >> n , n )
15 {
16 int i , j , k ;
17 for ( i = 0 ; i < n ; i ++ )
18 cin >> str [i];
19 memset(dis,0,sizeof(dis));
20 //求任意两车之间的距离
21 for ( i = 0 ; i < n - 1 ; i ++ )
22 {
23 for ( j = i+1 ; j < n ; j ++ )
24 {
25 for ( k = 0 ; k < 7 ; k ++ )
26 {
27 if(str[i][k]!=str[j][k])
28 dis[i][j]++;
29 }
30 dis[j][i]=dis[i][j];
31 }
32 }
33 //初始化集合和颠倒集合的距离
34 for ( i = 0 ; i < n ; i ++ )
35 lowdis[i]=dis[0][i];
36 int ans = 0 ; // 所求的解
37 for ( i = 0 ; i < n - 1 ; i ++ )
38 {
39 int mindis = ( 1 << 20 ) ;
40 for ( j = 0 ; j < n ; j ++ )
41 {
42 if( lowdis[j] && mindis > lowdis[j] )
43 {//寻找到集合距离最小的点
44 mindis = lowdis[j] ;
45 k = j;
46 }
47 }
48 ans += mindis ;
49 lowdis[k] = 0 ;
50 for ( j = 0 ; j < n ; j ++ )
51 {//更新各点到集合的距离
52 if(lowdis[j]>dis[k][j])
53 lowdis[j]=dis[k][j];
54 }
55 }
56 //输出答案，注意后面还有个‘.’
57 cout<<"The highest possible quality is 1/"<<ans<<"."<<endl;
58 }
59 return 0;
60 }